The air is a mixture of a number of gases. The maojr components are oxygen and nitrogen with approximate proportion of `20%` is to `79%` by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry's law constants for oxygen and nitrogen at 298 K are `3.30xx10^(7)` mm and `6.51xx10^(7)` mm respectively, calculate the composition of these gases in water.

Answer

`CHMi_(O_(2))=4.61xx10^(-5)`,`CHMi_(N_(2))=9.22xxxx10^(-5)`

Explanation

Air contains `20%` of `O_(2)` and `79%` of `N_(2)` by volume.
`P_(t otal)` of air in equilibrium with water `=10atm`
`:. p_(o_(2))=(20)/(100)xx10atm=2 atm=2xx760mm=1520mm`
`p_(N_(2))=(79)/(100)xx100atm=7.9atm=7.9xx760mm=6004mm`
Using Henry's law :
`p_(O_(2))=K_(H(O_(2)))xxCHMi_((O_(2)))`
or `CHMi_((O_(2)))=(p_(O_(2)))/(K_H(O_(2)))=(1520mm)/(3.30xx10^(7)mm)=4.61xx10^(-5)`
`p_(N_(2))=K_(H(N_(2)))xxCHMi_(N_(2))`
`:.CHMi_(N_(2))=(p_(N_(2)))/(K_(H(N_(2))))=(6004mm)/(6.51xx10^(7)mm)=9.22xx10^(-5)`